11

Source: p 77. Sweet Reason: A Field Guide to Modern Logic (2010 2 ed) by Henle, Garfield, Tymoczko.

I pursue only intuition; please do not answer with formal proofs or Truth Tables. I already comprehend, and so ask not about, the direct proof. Instead, why is this true intuitively? The citation below does not convince me intuitively, because it does not explain why intuitively, ¬(P ↔ Q) is not instead: ¬P ↔ Q or ¬P ↔ ¬Q.

The negation of a biconditional,

Harriet will go if and only if Gloria goes.        P ↔ Q

may be less familiar, but it still makes intuitive sense. It's another biconditional,

Harriet will go if and only if Gloria doesn't go.     P ↔ ¬Q

After all, P ↔ Q says P and Q have the same truth value. P ↔ ¬Q says that P and ¬Q have the same truth value, that is, that P and Q have different truth values.

  • 4
    Didn't you just answer your own question (intuitively, that is)??? If the semantics of P<==>Q is "P and Q do have the same truth value", then the semantics of .not.(P<==>Q) would intuitively just be "P and Q don't have the same truth value". In other words, the negation of "A satisfies B" is "A doesn't satisfy B". [Question of my own: how'd you get the math symbols without latex/mathjax?] – John Forkosh Jul 31 '16 at 6:11
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    What do you consider intuitive in logic? Can you explain intuitively why P=>(Q=>R) is equivalent to (P&Q)=>R so that we get an idea of what kind of explanation you want? – Colin McLarty Jul 31 '16 at 9:11
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    As far as I can see, this equivalence does not hold intuitionistically, so arguably doesn't make sense (except in special cases where P and Q are decidable, which one could argue for your example). – James Wood Jul 31 '16 at 10:50
  • It's not quite the same question, but you may be interested in math.stackexchange.com/questions/1870171/… – Mark S. Jul 31 '16 at 15:35
  • It is equivalent to ¬P ⟺ Q . – immibis Aug 1 '16 at 3:16
3

As you may know, the logical or (disjunction) is inclusive. That is, P ∨ Q means that either P is true, Q is true, or both. But usually in natural language we don't mean that when we use 'or'; we usually mean it to be exclusive, i.e. to rule out the case where both P and Q are true (e.g. "I want either coffee or tea").

Exclusive or is translated as ¬(P ↔ Q). It reads: 'either P or Q, but not both', or 'P and Q exclude each other: when one is true the other is false'. So it should explain intuitively why it is equivalent to P ↔ ¬Q. It is also equivalent to ¬P ↔ Q.

Example:

¬(P ↔ Q): the tea is either hot or cold, but not both.

P ↔ ¬Q: if the tea is hot, then it's not cold. And vice versa.

¬P ↔ Q: if the tea is not hot, then it's cold. And vice versa.

All three are equivalent.

  • I removed the parentheses because I fear that they may cause one to interpret the reverse implication as parenthetical or optional, but they are not because these are equivalences. – Greek - Area 51 Proposal Aug 2 '16 at 19:56
7

Intuitively, the statement P ↔ Q means that the truth values of P and Q are equal. Therefore, the negation of this statement entails their truth values must differ, i.e. P ↔ ¬Q.

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    What intuition is added by stating the definition? – jobermark Aug 1 '16 at 17:40
  • It's stating the definition differently, as to make it intuitively understandable (i.e. no conscious reasoning required). Intuition is relative. Complicated proofs can be explained in high-level terms that make it intuitively clear to a mathematician, but still convoluted to a layman. – xntric_ngneer Aug 2 '16 at 8:31
6

"Harriet will go if and only if Gloria goes" means the statements "Harriet will go" and "Gloria will go" have the same truth values; that is to say

either "both Harriet and Gloria will go" or "both Harriet and Gloria will stay"

Therefore the negation of "Harriet will go if and only if Gloria goes" is

"not both Harriet and Gloria will go" and "not both Harriet and Gloria will stay"

i.e.

"if Harriet goes then Gloria will stay" and "if Harriet stays then Gloria will go"

In other words: "Harriet will go if and only if Gloria stays".

3

I am not sure exactly what is missing here, intuitively, but maybe it is the fact that omitted variables are always free to take any value. Propositions must apply equally, independent of any circumstances not mentioned within the propositions themselves.

Given that convention, "A ↔ B" intuitively means "A and B are equivalent in meaning (under all circumstances)" because whenever one is true, so is the other and whenever one is false, so is the other.

The opposite of that fact, ¬(A ↔ B) does not just mean they are not equivalent, it means they are never equivalent, in any circumstance. That can only happen if they are opposite statements – one of them being true makes the other false. Otherwise, they could occasionally be equally true due to some surrounding circumstance that creates an unforeseen relationship between their referents.

In that case, the one really does mean the exact opposite of the other, which (using the same interpretation we started with) is what we mean by A ↔ ¬B.

(Note that your first possible alternative is just the same, ¬P ↔ Q means the same thing as P ↔ ¬Q. There is only one way for statements to be exactly opposite, it is a symmetrical relationship. And your other possible alternative is equivalent to the starting point. When the opposites of two statements say the same thing, the original two statements must have said the same thing. So neither is really an alternative.)

0

Lets start with (P<->Q) =/= P <-> ~Q.
Since the right side is false when the left side is true, then how do I make the left side false? Negate it, and obtain ~(P<->Q)!

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    This is arguing formally and ignoring a (as stated) perfectly apprehended proof already given in the question. I do not see how this is giving the intuition asked for. – Philip Klöcking Aug 2 '16 at 20:04

protected by Philip Klöcking Aug 2 '16 at 20:04

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